Interview question: Realize the arithmetic of addition, subtraction, multiplication and division of large numbers

#include <iostream>
#include <cstdlib>
#include <vector>
#include <sstream>
#include <algorithm>
using namespace std;

class Solution {
public:
string addTwoString (string nums1, string nums2) {
int n1 = nums1.size ();
int n2 = nums2.size ();
// Easy to operate, unified conversion to long number plus short number
if (n1 < n2)
return addTwoString (nums2, nums1);

int minlen = n2;
int maxlen = n1;
string result;
int carry = 0;
for (int i = 0; i < minlen; i++) {
int sum = nums1[--n1] - '0' + nums2[--n2] - '0' + carry;
carry = sum / 10;
int a = sum % 10;
result.push_back(a + '0');
}
for (int i = minlen; i < maxlen; i++) {
int sum = nums1[--n1] - '0' + carry;
carry = sum / 10;
int a = sum % 10;
result.push_back(a + '0');
}

if (carry != 0)
result.push_back(carry + '0');

reverse(result.begin(), result.end());
return result;
}

string multiplyTwoString(string nums1, string nums2) {
string result = "0";
if (nums1 == "0" || nums2 == "0")
return result;

vector<string> tmpResult;
int n1 = nums1.size();
int n2 = nums2.size();

int carry = 0;
reverse(nums1.begin(), nums1.end());
reverse(nums2.begin(), nums2.end());

for (int i = 0; i < n1; i++) {
string str;
for (int j = 0; j < n2; j++) {
int sum = (nums1[i] - '0') * (nums2[j] - '0') + carry;
carry = sum / 10;
int a = sum - carry * 10;
str.push_back(a + '0');
}
if (carry != 0)
str.push_back(carry + '0');
carry = 0;
reverse(str.begin(), str.end());
for (int k = 0; k < i; k++) {
str.push_back('0');
}
tmpResult.push_back(str);
}

int size = tmpResult.size();

for (int i = 0; i < size; i++) {
result = addTwoString(result, tmpResult[i]);
}
return result;
}

string substractTwoString(string nums1, string nums2) {
string result;
int n1 = nums1.size();
int n2 = nums2.size();

// judge the sign is positive and negative
char sign = '+';
if (n1 <n2) {
sign = '-';
nums1.swap (nums2);

} else if (n1 == n2) {
for (int i = 0; i < n1; i++) {
if (nums1[i] > nums2[i]) {
break;
} else if (nums1[i] < nums2[i]) {
sign = '-';
nums1.swap(nums2);
break;
}
}
}
int borrow = 0;
reverse(nums1.begin(), nums1.end());
reverse(nums2.begin(), nums2.end());

n1 = nums1.size();
n2 = nums2.size();

for (int i = 0; i < n2; i++) {
int r = nums1[i] - nums2[i] - borrow;
borrow = 0;
if (r < 0) {
r = r + 10;
borrow = 1;
}
result.push_back(r + '0');
}

for (int i = n2; i < n1; i++) {
int r = nums1[i] - '0' - borrow;
borrow = 0;
if (r < 0) {
r = r + 10;
borrow = 1;
}
result.push_back(r + '0');
}

for (int i = n1 - 1; i >= 0; i--) {
if (result[i] == '0')
result.erase(result.begin() + i);
else
break;
}
reverse(result.begin(), result.end());
if (sign == '-') {
result.insert(result.begin(), '-');
}
return result;
}

string divideTwoString(string nums1, string nums2) {
string result;
result = substractTwoString(nums1, nums2);
if (result[0] == '-') {
result = "0";
return result;
}
result = "";
int n1 = nums1.size();

stringstream stream;
string dividend;

for (int j = 0; j <n1; j ++) {
dividend.push_back (nums1 [j]); // Dividend
// from 9 to 0, tentatively looking for a quotient, when the remainder is not negative, it is the current bit values
for (int I =. 9; I> = 0; i--) {
stream.str ( "");
Stream << I;
String stream.str S = ();
String = multiplyTwoString SJ (nums2, S) ;
string remain = substractTwoString (dividend, sj);

if(remain[0] != '-'){
result.push_back(i+'0');
dividend = remain;
break;
}
}
}
//去掉商前面的零位
int size = result.size();
for(int i=0;i<size;i++){
if(result[0]=='0'){
result.erase(result.begin());
}else {
break;
}
}
//四舍五入
string s = substractTwoString(multiplyTwoString(dividend,"2"),nums2);
if(s[0] == '-') addTwoString(result,"1");

return result;
}
};

int main() {
string nums1 = "123456";
string nums2 = "789";
Solution s;
cout << s.addTwoString(nums1, nums2) << endl;
cout << s.multiplyTwoString(nums1, nums2) << endl;
cout << s.substractTwoString(nums1, nums2) << endl;
cout << s.divideTwoString(nums1, nums2) << endl;
return 0;
}